Question

A short bar magnet of magnetic moment 2.5 Am\(^2\) is kept in a uniform magnetic field of \(4 \times 10^{-5}\) T. The work done in moving the magnet from its most stable position to most unstable position is

• A. \(40 \times 10^{-5}\) J
• B. \(25 \times 10^{-5}\) J
• C. \(10 \times 10^{-5}\) J
• D. \(20 \times 10^{-5}\) J

Hint:

Remember the key positions for a dipole in a field: - Most Stable: \(\theta=0^\circ\), aligned with the field, minimum potential energy (-MB). - Most Unstable: \(\theta=180^\circ\), anti-aligned with the field, maximum potential energy (+MB). - Zero Potential Energy (by convention): \(\theta=90^\circ\), perpendicular to the field.

Answer 1

The Correct Option is D